Asked by Sarah
A naked person with a surface area of 1.55 meters squared and a skin temperature of 37.3 degrees is in a sauna room at T = 82.0 degrees. The person's skin has an emissivity of 0.910. She evaporates sweat from her body to balance the rate of heat absorption by radiation.
(i) Neglecting any heat production due to metabolism, what is the net rate of heat absorption by the person due to radiation? in W
(ii) How much sweat must be evaporated per hour to balance a net rate of heat absorption of H = 0.850 × 103 W? in kg/hr
(i) Neglecting any heat production due to metabolism, what is the net rate of heat absorption by the person due to radiation? in W
(ii) How much sweat must be evaporated per hour to balance a net rate of heat absorption of H = 0.850 × 103 W? in kg/hr
Answers
Answered by
drwls
Radiation power absorbed
= Area*sigma*emissivity*(T2^4 - T1^4)
T1 and T2 are the skin and sauna wall temperatures, respectively, in Kelvin.
sigma is the Stefan-Boltzmann constant. (Look it up).
ii) Divide the heat absorption rate (W or J/s) by the heat of vaporization of water, 2.26*10^6 J/kg. Then convert kg/s to kg/hr
= Area*sigma*emissivity*(T2^4 - T1^4)
T1 and T2 are the skin and sauna wall temperatures, respectively, in Kelvin.
sigma is the Stefan-Boltzmann constant. (Look it up).
ii) Divide the heat absorption rate (W or J/s) by the heat of vaporization of water, 2.26*10^6 J/kg. Then convert kg/s to kg/hr
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.